Question

If the integral $I=\displaystyle \int \frac{2 x^{2}}{4 + x^{2}}dx=$ $2x-f\left(x\right)+c$ , where $f\left(2\right)=\pi $ , then the minimum value of $y=f\left(x\right)\forall x\in \left[- 2,2\right]$ is (where, $c$ is the constant of integration)

• A. $0$
• B. $-\pi $
• C. $2\pi $
• D. $-4\pi $

Answer 1

Answer: (B)

$I=2 \int\left(\frac{x^{2}+4-4}{x^{2}+4}\right) d x$
$=2 \int\left(1-\frac{4}{x^{2}+4}\right) d x$
$=2\left(x-\frac{4}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right)+c$
$=2 x-4 \tan ^{-1}\left(\frac{x}{2}\right)+c$
$\therefore f(x)=4 \tan ^{-1}\left(\frac{x}{2}\right)$
So, $\min (f(x))=4 \tan ^{-1}(-1)=-\pi$